TPS61088: TPS61088 EN pin and load disconnection

Although I dont know if load disconnection is a must, we used TPS62180 before for similar application and it seemed to work quite fine, I mean in terms of saving power, and we thought it was disconnecting the load but maybe it works the same as TPS61088. Basically we just need to save power when we shut this off, it is used to power the motor driver DRV8305 also from TI. Could you tell me if in this case, load disconnection is necessary or not please?
TPS61178 has the load disconnection feature, but the maximum switching current is typically 8A. Since we are boosting up the voltage from 3.7V to 9V with 1-3A driving capacity, would that be a problem?
I saw in TPS61178, there is a P-MOSFET added to control the load connection, could we use the similar circuit for TPS61088, and control the MOSFETs by uC? Do you have some typical schematics for that?
The project, I am afraid I could not reveal too much info without a signed NDA since it is strictly confidential, I do apologize for that. But we do have a lot of TI devices used in this project. It is a handheld medical devices, and we are Ferrosan Medical Devices in Denmark, the yearly volume for the device would be 2-3k, it will ramp up the next two years, and drop down after that. The market will be saturated after 5 years, but then we will have newer version come into the market.

Best regards
Weinan Ji

Hi Helen,

Thanks for your quick reply.

I am just wondering if it is really needed here in our application. Also as I mentioned we used TPS62180, which is a buck converter, and we thought we were disconnect the load because of the misunderstanding. Could you verify that TPS62180 does not have load disconnection feature either please?

Thanks.

No no, that was another application, sorry for the confusing topic. But then I think load disconnect is necessary, but I am not really understanding the application note you sent to me. Is that problem solved by just adding a P-MOSFET?
Thank you.

BR
Weinan

Thank Helen, I think I got the point. I am working on the design at this moment, but the loop compensation is really confusing me, it starts on the datasheet from Page17 to Page19.
Questions:
1. I think basically equation 13 to 17 are the modeling of the compensation network, but I do want to know what is fRHPZ. I know we could calculate this value by equation 16, but what is that?
2. D is the switching duty cycle, but I want to know what value should I plug in for the calculation?
3. How could I design this loop crossover frequency fc?
4. Is output capacitor Co C9+C4 on Page13?

Thank you in advance.

Best regards
Weinan Ji

Hi Helen,

Is CSD25304W1015 OK?

Hi Wennan,

fRHPZ is the riht half plane zero of the boost converter.

D can be calculated by equation D=1-(Vin/Vo)

For the boost converter, you can choose  fc=5k-10k.

Yes, Co=C9+C4. But you should use the effective capacitance, that is for a 22uF/25V/1206  caramic, the effective capacitance maybe only 10uF. You need to check  you datasheet.

Since your input and output is similar to our EVM board,  you can use the same compensation value as what is on the EVM board.

Hi Helen,

Thanks for your reply.

Since my motor terminal resistance is 1.8 ohm. I guess I could use this value as Ro (output load resistance), right?

Let me do this calculation again, so you could correct me.

If 1.2uH inductor is chosen.

fRHPZ = (1.8*(1-3.3/9)^2)/(2*pi*1.2u) = 94.8kHz
fSW = 750kHz this is what we fix

Then the principle is to set the crossover frequency no higher than the lower of either 1/10 of switching frequency fSW, or 1/5 of the RHPZ frequency.
1/10 fSW = 75kHz
1/5 fRHPZ = 18.96kHz

Thus if the crossover frequency fc is less than 18.96kHz, it will be ok, right?

And since you were suggesting from 5kHz to 10kHz, which is also in the range. I wonder what is the best trade-off crossover frequency, the higher the better or lower, and why?

With Equation 18 in the datasheet, we could calculate R5. If we say fc = 10kHz, and Co is 4 such 22uF capacitors with 1 such 1uF capacitor in parallel so that Co = 10uF*4 = 40uF (effective capacitance approximately) then
R5 = (2*pi*9*0.08*10k*40u)/((1-3.3/9)*1.212*190u) = 12.47k, as I am using PFM mode, the reference voltage will be 1.212V according to the electrical parameter table.

Then we calculate C5 with Equation 19,
C5 = (1.8*40u)/(2*R5) = 2.8869nF = 2887pF

Last thing is the C8 since we have Equation 20, we have 4 of this 22uF capacitors(ESR is approximately 30m Ohm), I think in pararllel the total equivalant resistance would be 30m/4 = 7.5m Ohm, if I am not right please correct me. So we get,
C8 = 7.5m*40u/12.47k = 24pF

Could you confirm if my calculation is right please?

Thank you and sorry for the long calculation.

Best regards
Weinan

Usually we make the cross over frequency smaller than the calculated value, so it be better choose fc=10kHz.

You can connnect a Vo=9V DC source to you mortor to check the output current, R=9/Isource.

Hi Helen,

OK, I will fix fc = 10kHz

Then do you mean that the output resistance Ro = 9V/3A = 3 ohm, because that is the setup?

Besides that, the rest of my calculation is correct right?

Thanks.

Best regards
Weinan

Yes。 3A is the load current.

The rest of the calculation is right!